Đáp án:
\(1 - \dfrac{{{{\left( {2x - 1} \right)}^2}\left( {2x + 1} \right)}}{{x{{\left( {x - 1} \right)}^3}\left( {{x^2} + x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
U = 1 - \left[ {\dfrac{{2{x^2} + x - 1}}{{\left( {1 - x} \right)\left( {x + 1} \right)}} + \dfrac{{2{x^3} + {x^2} - x}}{{\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}} \right]:\dfrac{{ - \left( {x - 1} \right)x\left( {x - 1} \right)}}{{2x - 1}}\\
= 1 - \left[ { - \dfrac{{\left( {2{x^2} + x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{{\left( {2{x^3} + {x^2} - x} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}} \right].\dfrac{{2x - 1}}{{ - x{{\left( {x - 1} \right)}^2}}}\\
= 1 - \left[ {\dfrac{{ - 2{x^4} - 2{x^3} - 2{x^2} - {x^3} - {x^2} - x + {x^2} + x + 1 + 2{x^4} - 2{x^3} + {x^3} - {x^2} - {x^2} + x}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}} \right].\dfrac{{2x - 1}}{{ - x{{\left( {x - 1} \right)}^2}}}\\
= 1 - \dfrac{{ - 4{x^3} - 4{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{2x - 1}}{{ - x{{\left( {x - 1} \right)}^2}}}\\
= 1 - \dfrac{{\left( {1 - 2x} \right)\left( {2x + 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{2x - 1}}{{ - x{{\left( {x - 1} \right)}^2}}}\\
= 1 - \dfrac{{\left( {1 - 2x} \right)\left( {2x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{2x - 1}}{{ - x{{\left( {x - 1} \right)}^2}}}\\
= 1 - \dfrac{{{{\left( {2x - 1} \right)}^2}\left( {2x + 1} \right)}}{{x{{\left( {x - 1} \right)}^3}\left( {{x^2} + x + 1} \right)}}
\end{array}\)
( bạn xem lại đề nhé, kết quả rất to nha )
